2-1 某双导线的直径为2mm ,间距为10cm ,周围介质为空气,求其特性阻抗。某同轴线的外导体内直径为23mm ,内导体外直径为10mm ,,求其特性阻抗;若在内外导体之间填充εr 为的介质,求其特性阻抗。
解:双导线:因为直径为d =2mm =2×10-3
m
间距为D =10cm =10-1
m 所以特性阻抗为
d
D d D d D Z 2ln 120]1)(ln[
12020≈-+= Ω=??=--6.552102102ln 1203
1
同轴线:因为外导体内直径为2b =23mm 内导体外直径为2a =10mm 当εr =1时 特性阻抗为Ω===5010
23ln 160ln
60
0a b Z r
ε 当εr =时 特性阻抗为Ω===3.3310
23
ln
25
.260ln
60
0a
b Z r
ε
2-2 某无耗线在空气中的单位长度电容为60pF/m ,求其特性阻抗和单位长度电感。
解法一:在空气中ε=ε0 、μ=μ0 、C 1=60pF/m
0011εμμε==?C L
所以
H C L 7121610011085.11060/1091
/---?=??==εμ
Ω=??==--6.5510
61085.111
7
110C L Z
解法二:在空气中8103?=p υ
所以Ω=???==
-6.5510
601031
112
810C Z p υ
H Z L p
7
8
11085.110
36.55-?=?=
=
υ 2-4 求内外导体直径分别为0.25cm 和0.75cm 空气同轴线的特性阻抗;在此同轴线内外导体之间填充聚四氟乙烯(ε0 =),求其特性阻抗与300MHz 时的波长。
解:因为内外导体直径分别为2a =0.25cm ,2b =0.75cm , 当在空气中时 ε0 =1
Ω===
9.6525
.075.0ln 160ln
60
0a b Z r
ε 当填充聚四氟乙烯时ε0 =
Ω===
5.4525
.075.0ln 1.260ln
60
a b Z
r
ε 因为0111
1εμεβωυr p C L ===
m f f
r p
69.01
.21120
==
=
=
=
εμευβ
π
λ
2-5 在长度为d 的无耗线上测得)(d Z sc
in
、)(d Z oc in 和接实际负载时的)(d Z in ,证明 )
()()
()()
(d Z d Z d Z d Z d Z Z oc
in
in in sc in oc
in L --=
假定Ω=100)(j d Z sc in
,Ω-=25)(j d Z oc in ,Ω?∠=3075)(d Z in ,求L Z 。 证明:对于无耗线而言d tg d jZ Z d
tg jZ d Z Z Z in in L
ββ)()(000
--= (1)
且)()(0d Z d Z Z oc
in sc in = (2)
d tg jZ d Z sc in β0)(= (3)
)
()
(d Z d Z d jtg oc in sc
in =β (4)
将(2)、(3)、(4)式代入(1)式中有)
()()
()()
(d Z d Z d Z d Z d Z Z oc
in
in in sc
in oc in L
--=
当Ω=100)(j d Z sc in ,Ω-=25)(j d Z oc
in ,Ω?∠=3075)(d Z in 时
)25(757510025
3030j e e j j Z j j L ----=??
?∠=+??+=
2.2255.622
37523
18755.1562j
j
2-6 在长度为d 的无耗线上测得Ω=50)(j d Z sc
in
,Ω-=50)(j d Z oc
in ,接实际负载时,VSWR =2,d min =0,λ/2, λ,…,
求Z L 。
解:因为VSWR =2,所以13
1
11||L <=+-=ΓVSWR VSWR ,因而为行驻波状态
所以)12(44min ++=
n d L λ
φπλ n=0,1,2…… 当n =0时,044=+λφπλL 所以得到πφ-=L ,n d 2
min λ
= n=0,1,2…… 所以3
131||L
L -==Γ=Γ-πφj j e e L Ω=-?==50)50(50)()(0j j d Z d Z Z oc
in sc in
3
1
505000L -=+-=+-=
ΓL L L L Z Z Z Z Z Z
所以 Ω=25L Z
2-7 设无耗线的特性阻抗为100Ω, 负载阻抗为50-j 50Ω, 试求ГL 、VSWR 及距负载λ处的输入阻抗。
解:0
L Z Z Z Z L L +-=
Γ=
=Γ-Γ+=
=|
|1|
|1L L VSWR ρ
=++=)
()
()(000
in d tg jZ Z d tg jZ Z Z d Z L L ββ+
2-10 长度为3λ/4,特性阻抗为600Ω的双导线,端接负
载阻抗300Ω;其输入端电压为600V 、试画出沿线电压、电流和阻抗的振幅分布图,并求其最大值和最小值。
解: 0
L Z Z Z Z L L +-=
Γ=-1/3=1/3exp(j π)
V
V V e e V V e e V d V L L j j L d j L d j L L 450600)3/4()3
1
1()4/3()
||1()()3(2/3)2(-==-=+=∴Γ+=++-+-Φ+πππββλ
|
)(/)(||)(|)]/2cos(3/29/10[450)]2cos(||2||1[|||)(|)]/2cos(3/29/10[450)]2cos(||2||1[|||)(|2/12/122
/12/12d I d V d Z d d V d I d d V d V in L L L L L L L L =+=-ΦΓ-Γ+=-=-ΦΓ+Γ+=++λπβλπβ
振幅|V(d)|、|I(d)|、|Zin(d)|随d 的变化图
Ω
==Ω===Γ-=
=Γ-==Γ+==Γ+=++++300|)(|/|)(||)(|1200|)(|/|)(||)(|5.0|]|1[|
||)(|300|]|1[|||)(|1|]|1[|
||)(|600|]|1[|||)(|max min min min max max 0
min min 0max max d I d V d Z d I d V d Z A Z V d I V V d V A
Z V d I V V d V in in L L
L L L L L L
2-11 试证明无耗传输线的负载阻抗为
1
min 1
min 0
L 1d jKtg d jtg K Z Z ββ--=
式中,K 为行波系数,d min1为第一个电压驻波最小点至负载的距离。
证明:因为d
th Z Z d
th Z Z Z d Z L L γγ++=000
in )(
对于无耗线d jtg d j th d th j ββγβγα====)(,,0
则得到d
tg jZ Z d
tg jZ Z Z d Z L L ββ++=000
in )(
所以可以得到d
tg d jZ Z d tg jZ d Z Z Z in in ββ)()(000
L --=
又因为当电压最小点时,电流为最大点,即
|]|1)[()(1min min 1min L d V d V Γ-=+
|]|1)[()(1min max 1min L d I d I Γ+=+
所以K Z d I d V d I d V d Z L L 01min 1min 1min 1min in |]
|1)[(|]
|1)[()()()(=Γ+Γ-==++
1min 001
min 000
L d Ktg jZ Z d tg jZ K Z Z Z ββ--=
1
min 1
min 0
L 1d jKtg d jtg K Z Z ββ--=
2—12画出图2—1所示电路沿线电压、电流和阻抗的振幅分布图,并求其最大值和最小值。
(图)
解:首先在BC 段,由于Z 0=Z 01=600Ω,Z L =400Ω 且因为d =λ/4
所以在BB ’处向右看去,Z in =Z 012/Z L =6002
/400=900Ω
又由于BB ’处有一处负载R =900Ω,所以对AB 段的传输线来说 终端负载为Z L ’=
Z in 0450
450450
450''0202=+-=+-=
ΓZ Z Z Z L L L 0)(=Γ?d z
j z j z j e V e d e V z V βββ-+-+=Γ+=00])([)(z j z
j z
j z
j e Z V e
I e
d e I z I ββββ-+-+-+==Γ-=0
00
])([)(Ω
==Γ-Γ+=450)
(1)
(1)(00
Z d d Z z Z in V V in 450450
450450
900
=+=1A z j e z V β-=450)(z j e z I β-=)(Ω
==450)(0Z z Z in d
j d L L L e Z Z Z Z β2005
1
51600400600400--=Γ?-=+-=+-=
Γ1
||<ΓL ]
1[)(2d j L d j L e e V d V ββ-+Γ+=]
1[)(2d j L d
j L e e I d I ββ--Γ-=V
V V
e e V l V L j L j L 375||450|]1[||)(|4/24/=?=Γ+=+-+λβλβV V V L L 4505
6
375|]|1[||||max =?
=Γ+=+V V V L L 30054
375|]|1[||||min =?
=Γ-=+V
Z V I L L L 16/15400/375/===++)
(8
9561615|]|1[||||max A I I L L =?=Γ+=+
)(4
3
541615|]|1[||||min A I I L L =?=
Γ-=+
d
j d
j in e e Z d d Z d Z ββ220
05
1151
1)
(1)(1)(--+-=Γ-Γ+=....2,1,0242424max =+=+=+=
n n n n d L λλλππλλφπλ....2,1,02
24244)12(4min
=+=++=++=n n n n d L λλλλλλφπλ2
12
121//in in in in in in in Z Z Z Z Z Z Z +?==d
tg jZ Z sc in β0=d
tg jZ Z in β01=d
tg jZ d tg jZ d tg jZ Z in ββπλ
β0002)()2
(-=-=-=0
21=+in in Z Z ∞
→in Z 2
12121//in in in in in in in Z Z Z Z Z Z Z +?=
=d
ctg jZ Z oc in β0-=d
ctg jZ Z in β01-=d
ctg jZ d ctg jZ d ctg jZ Z in ββπλ
β0002)()2
(=--=--=0
21=+in in Z Z ∞
→in Z Ω
=--=∴160)4/()4/()
4/()4/(000L βλλβλλtg jZ Z tg jZ Z Z Z in in =
=L
L Z L V P 2
|)(|211’1’
7100=+-=
ΓZ Z Z Z in in 3/4||1|
|1=Γ-Γ+=VSWR 1’21220
0000000j Z Z j Z Z Z j Z Z Z Z Z in in
+=++-+=+-=Γ2/2223||1||1+=Γ-Γ+=
VSWR 1’0000000=+-=+-=ΓZ Z Z Z Z Z Z Z in in 1|
|1||1=Γ-Γ+=VSWR 1
’
5
3
44000000=
+-=+-=
ΓZ Z Z Z Z Z Z Z in in 4|
|1|
|1=Γ-Γ+=
VSWR 8cm1.5cm 16cm
min
min 0
L 1d jtg VSWR d jVSWRtg Z Z ββ--=λ
π
β2=
1.5cm
668.05.116
2min =?=π
βtg
d tg ?
-=--=71.34L 57.39668.02336.1150j e
j j Z 31||||00=+-=ΓZ Z Z Z L L L 0
2||21Z V P L ++=02||P P L Γ=-9
1
||20=Γ=-L P P 9
8
||11200=Γ-=-=-+L P P P P )
(54.931
lg 20||lg 20lg 10dB P P Lr L =-=Γ-==-+)
(51.089
lg 10||11lg
102
dB L L R ==Γ-=15cm
28.4
0015090750.481509075j L L L Z Z j e Z Z j -+-Γ===+++2d in j L e βΓ=Γ-88
2f 26104310ππβπυ???===6
28.4
(28.4216)187.6
5
in 0.480.48
0.48j j j j e
e
e e π--Γ===-max (1
)0.48)20.9()
L L M V E V +=+Γ=+=14.8()V
V =?
220.03450.0690.480.480.480.930.45al in L e e e --?-Γ=Γ=?==?=20lg r L =-Γ620
110
0.5
-Γ=≈2020
210
0.1
-
Γ==111110.5
3110.5
VSWR +Γ+=
==-Γ-222110.1
1.22110.1VSWR +Γ+=
==-Γ-30cm G Z Z
=,,2(8.686)r i r L L L al =+30cm 2
Γ
,2
110lg
20lg 210lg513.01r L L dB ==+=Γ
,,2(8.686)
r i r L L L al =+???
12
1610lg
r L dB ==Γ
2
2
13lg
0.2515
=
?Γ=Γ
12
110lg
1.261R L dB
∴=≈-Γ
22
12010lg
r L dB ==Γ
2
0.01
Γ=22
110lg
0.0441R L dB =≈-Γ
L
Z H
μ0
Z 0100200
2450L L Z j Z j Z +=
==+L Z 1in Z 1in Z 1in y 1in y '0Z j L ω1in y τ2in y 28 2.0in y j =-2in y 2in y '
j c ωτ?
?
6
10??
12
10-?
21
0.1250.002
80.115
in Z j j '∴=
=+-220 6.250.09
in in Z Z Z j ''∴==+2020
6.250.095043.750.09
0.778
6.250.095056.250.09in in Z Z j j j j Z Z τ'-+--+====+++'+11ττ+-10.77810.778+-15cm80cm min 316d λ=
1
VSWR
min r min
r 3
16
λL Z L Z 0Z L Z L Z ?L
Z L Z L Z 1 1.42 1.420.230.34(1.42)(1.42) 1.964
L L j
j y j Z
j j ++=
===+-++L
Z L
Z L
Z 0170.7Z ===Ω7.5cm
0100250
L L Z Z Z =
==9
10
0170.7Z ===Ω
8
9
3100.310m
f υ
λ?===0.075m
0.1
τ<1
0.1
1
VSWR VSWR τ-=
<+0/4λ01Z 0/4λ050Z =Ω
200L Z =
Ω
01100Z ===Ω893100.122.510m f υ
λ?==
=?1
0.120.0344
l m λ==?=1f 11
5l λ=012L L Z Z Z ==L
Z 11
5
λ1in Z 1in Z 1011in in Z Z Z =010in
in in Z Z Z Z -Γ=+20cm
863103
0.2360.2360.0354********
f υ
?=?=?=?in
Γ2210/50/20125L L Z Z Z ===Ω
1102/L L Z Z Z =1
L Z in Z in Z 000()L in
L Z jZ tg d
Z d Z Z jZ tg d
ββ+'=+λ2π
βλ
=
(10075)500.24()50476250(10075)0.24in j j tg Z d j j j tg π
π
++'∴=?≈-++0()0.95 1.25in
Z d j Z '≈-0Z ()in Z d '0
in Z Z ∴=Ω
in
Z 0
in Z jx Z +=050in Z Z jx jx
=-=-150
in x
Z j
=-02530
50L L Z j Z Z +==01Z 01Z 0.23686
31030.2360.2360.0354********f υ?=?=?=? 3.54cm ω
ω
99
6
55
555510 4.381020001024ω
ππ
--=
=?=???0Z 01
Z Ω
L
Z 30
Ω
f 010.50.6L L Z Z j Z ==+0.10L d λ=1d λ1L d d -λ8
9
3100.064210???0.0096m in
Z 9011
22210 1.1250
C fxZ ππ-∴=
=
????in Z 1f 1λ1λ1
λλ
L Z 1λ1in
Z '911011122 1.81050f x Z x ππ-=????111in in Z Z jx '=+1in Z Ω101
10510
9510
in in in Z Z j Z Z j ---Γ==+-111
11in in VSWR +Γ=
-Γ2f 2λ2λ2λ2λL Z 2λ2in
Z '9220211
22 2.21050f x Z x ππ-=????2x 22in in Z Z jx '∴=+2
in Z 2
in Z 202205 6.5
105 6.5
in in in Z Z j Z Z j -+Γ=
=
++222110.078110.078
in in VSWR +Γ+=
=
-Γ-0Z ΩL Z L Z ΩL Z Ω1509075L j Z -=15090
75L j Z +=0Z
ΩL Z ΩΩΩ01Z λ01
Z min
0min
1L j tg d Z Z j tg d ρβρρβ-=-ρ21240305022j tg
d
j jtg d
π
λ
πλ-∴-=?
-2tg
d
π
λ
∴18d
λ
=
8d λ∴=144030504j tg
j jtg π
ρπρ-+=?
-12ρ∴=1460504
j tg
jtg π
ρπρ-=?-6
5
ρ∴=
L Z 00.80.6L L Z Z j
Z ∴==-0
in Z Z ρ
=
ρ∴↑in Z ∴↓0110Z =ΩL Z ΩΩL Z L
y L y L
τ1g 1y 1y 'λλ2y 2y 'L y =∞λλλl 'λλλ9
7.510?8
8
3107.510f υ
λ?==?0.04m4cm0
.076cm0.472cm 9
610?L y 1y 1y '0()tan sc in Z d jZ d β=cot sc
in y j d β∴=-1cot 0.9
d β∴=11cot (0.9)
d β
-=
L y L
y '2cot d β12cot (0.5)
d β
-=ΩΩ80cm 0300250
500
L L Z j Z Z +=
=ρmin d λ
λ
12.48cm 1Z ==?Ω12.48cm ΩL Z 0d L Z 0d λL Z L Z ρ
2π
βλ
=
min 0.208d λ=min 0
min 11 2.4(20.208)
500 2.4(20.208)
L j tg d j tg Z Z jtg d jtg ρβπρβπ--?==?
--?1 2.4 3.7500 2.4 3.70j j -??
-?18.88
500 2.4 3.70
j j -≈?-?Ωmax
V
min V min1d λL Z max
V
min
V max min
200
540
V VSWR V
=
=
=15121513L VSWR VSWR --Γ=
==++L Z 0L L Z Z Z ∴=L y L y τ1g 1y 2y λλλλλλ2y 2
y 'L y =∞2y 2y 'λλλl 'λλλ0Z ΩL Z Ω01Z Ω0
L
L Z Z Z =λL Z 1Z λ1Z 'λλλλd 'λλλλ2Z 32001/Z Z Z Z =2
Z '32001/Z Z Z Z ''=3Z λZ 'λλλ3Z 3Z 'λl 'λλλλλ2d λ020sin L Y G d β≤≤
2sin sin 0.588105
πλπ
λ==L y 0
Y 00 2.89L G Y ≤≤L Y 'ΩΩλλ1l 2l 300300
600
L
j Z +'=L
y 'λL y 'ΓλL y λL y L g 1y 1jb ∴1l ∴λλ1y Γλ2y ∴2 1.2jb j ∴=-2l ∴λλΩ1l λ2l λL Z Ω10in
Z Z '=21in Z '∴=2in
y '∴2in y ?2in y '2in y ∴2in y τλ1in y '1in y '1in y '1in y ?1in y ∴0Z ΩΩ3cm2.5cm20cm 1l 2l 3l
00.6L L
Z Z j Z '==-1d λ8
d λ
?=L
Z '0180L y 'L y 'λL y 'λL y L y λL y L g 1y 1jb λ1l ∴λλλ1y λ2jb λ2l ∴λλ2y λ3y 3y 3jb λ3l ∴λλλλλλ0Z ΩλL Z L
Y 1
L
Z Y λλλD Y 2Y 2Y 1jB 1.8m Ωr
εΩΩλλΩ0),(02=?υμt t
E 0),(0≡??υμt t E
8cm
4cm8cm4cm
GHz
f 3=GHz b
n a m f c 3)()(
212
2<+=
ππμε
π40010312103)()(2
89
22=??? ?
?????<+b n a m 56
.2422<+n m GHz
f 5=9/1010312105)()(42
8922=??
? ??????<+b n a m 2.7422<+n m 10.2mm0.0229m0.0102m
2
222)2()()(
c
f b n a m c πππκ=+=2
)()(22
2b n
a m c c f c c +==πκ10
f2a 01
f 20
f
GHz
f c 0.162
)0102.01
()0229.01(
3.02
2=+?=
0.0366m 01c 0.0609m0.0242m 01c 0.029
m6cm10.9cm10cm72.14mm34.04mm
22)()(2b
n a m c +=
λcm
a cTE 43.14210==λcm
b cTE 808.6201==λcm
b
a cTE 143.6)1
()1(22211=+=
λ10.9cm10.9cm14.43cm
cm
c
g 8.21)(
12
=-=
λλλ
λ12cm10cm14.43cm
cm
c
g 87.13)(
12
=-=
λλλ
λ
s m c
p /1016.4)(
182
?=-=
λλυ
υs m c
g /1016.2)(182?=-=λλυυ10TE 10
TE
10sin
j z
y j a
x
E H e a
βωμππ
--=
10sin
j z
x j a
x
H H e a
ββππ
-=
10cos
j z
z x
H H e a
βπ-=x z y E E E =
=g v =10
TE 10gTE v =mm10.16mm2a45.72mm3cm
cm
c
g 98.3)(
12
=-=
λλλ
λ12222
2158])()[(
)2(
-=+-=-=cm b
n a m C π
πλ
π
κκβΩ=-=64.497)2(1210a
Z TE λ
η
Ω=-=
2.221)
2(12
10a
a b Z e λ
η
2a4.572cm9
.144cm 2
)(
1c
g λλλ
λ-=
20.16mm2cm cm c
g 224.2)(
12
=-=λλλ
λ 1.5cm3cm4cm2.286cm1.016cm
22)()(2b
n a m cmn +=
λcm c 57.410=λcm
c 03.201=λcm c 856.111=λcm c 519.121=λcm c 99.012=λcm c 286.220=λcm
5.10=λc λλ<0cm 30=λc λλ<0cm 40=λc λλ<0cm c 57.410=λcm
c 286.220=λcm
cm 57.4286.20<<λGHz
f GHz 123.13562.60<<22)()(2b
n
a m cTEmn +=
λa cTE 210=λa cTE 210=λa cTE 210=λ5cm7cm6cm3cm7cm0.05
m0.025m a
u k mn
cmn =mn
cmn u a
πλ2=
m a cTE 0853.041.311==λm a cTE 041.064.101==λm a cTM 0655.062.201==λ7cm
01
0111cTM cTE cTE λλλλλλ>><6cm 01
0111cTM cTE cTE λλλλλλ<><3cm
01
0111cTM cTE cTE λλλλλλ<<<7cm
m
CTE g 125.0)
/(12
11=-=
λλλ
λr ε 1.016cm r
εr
ε3
10-=δtg g λp υd
α220min )()(
21
5b
n a m f GHz f r
c ππεμεπ+=
>=20
2022r r cTE a c
a c
f επεπ=
=01
0122r r cTE b c b
c f επ
επ=
=
10
102r cTE a c f ε=
01
20cTE cTE f f <10
205cTE cTE f GHz f f >=>10
20
25r r a c GHz a c
εε>
>10
82920
21
10310286.21051
r r εε>
????>-10
20
21381.01
r r εε>
>r
r εε>>6247.22620
.11456.1< 10)2/(1a gTE λλ λ-= cm f c r 425 .21051039 8=???==ελcm gTE 26.8) 04572.0/04.0(104 .02 10=-= λ2 10) 2/(1a pTE λυ υ-= s m c r /10225 .21038 8?=?= = ευs m pTE /1013.4) 04572.0/04.0(11028 2 8 10?=-?= υ)/(78.9)/(324.02tan 22tan 2 22 2m dB m Np a k d ==?? ? ??-??? ???? ? ??==πλπδλπβδα6cm0.06m0.03m a a u k cTE 841.1' 1111 ==cm a k CTE cTE 23.10841 .12211 11== = ππλGHz c f cTE c 93.211 === λm f c 1.010310398 00=??==λm CTE g 474.0) /(121100=-=λλλλΩ=== 1.1914g w f Z ωλβ ωμ 2cm0.01m m f c 03.0/==λm a u a TE cTE 0341.041.3/21111===πλm a u a TE cTE 0164.064.1/20101===πλm a u a TM cTM 0262.062.2/20101===πλ01